Question

1) Forty percent of the english students at Parkway High school had Dr.Mima for history, and the rest had Dr. P. Thirty percent of Dr.Mima's students earn an A in probability, while only ten percent of Dr. P's students earn an A in probability.

a) What percent of Parkway High School history students earn an A?

A= have Dr.Mima=0.40

B= rest have Dr.P=0.60

C= Dr.Mima students earn A = 0.30

C= Dr.P students earn A = 0.10



P(C)=P(A)P(C|A) + P(B)P(C|B)

= (0.40)(0.30) + (0.60)(10)

P(C) = 0.19 -answer



b) Tiffany did not earn an A in english. What is the probability that Tiffany had Dr.P for history?

Answer 1

a) What percent of Parkway High School history students earn an A?

A= have Dr.Mima=0.40

B= rest have Dr.P=0.60

C= student earns A

$P(C|A)=0.3$

$P(C|B)=0.1$

Total probability

$$P(C)=P(C|A)P(A)+P(C|B)P(B)$$ $$P(C)=0.3(0.4)+0.6(0.1)=0.18$$

18% of Parkway High School history students earn an A.

b) Tiffany did not earn an A in probability. What is the probability that Tiffany had Dr.P for history?

Let C be the event that student earns an A in probability. Then

$$P(\bar{C}|A)=1-0.3=0.7$$

$$P(\bar{C}|B)=1-0.1=0.9$$

Use the Bayes' rule

$$P(B|\bar{C})={P(\bar{C}|B)P(B) \over P(\bar{C}|B)P(B)+P(\bar{C}|A)P(A)}$$

$$P(B|\bar{C})={0.9(0.6) \over 0.9(0.6)+0.7(0.4)}={27 \over 41}\approx0.66$$

The probability that Tiffany had Dr.P for history is ${27 \over 41}\approx0.66$